Question

The shortest distance between the z- axis and the line $x + y + 2z - 3 = 0$, $2x + 3y + z - 4 = 0$, is
A) ${\text{1}}$
B) ${\text{2}}$
C) ${\text{3}}$
D) $4$

Answer 1

First calculate any two points that lie on the given line. And using that both points find the equation of the line, and also write the equation of the line for the z-axis. Then use the formula of shortest distance formula as $d = \dfrac{{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )}}{{\left| {(\overrightarrow b \times \overrightarrow d )} \right|}}$. Hence calculate each and every vector and then, put it in the above formula and our required answer will be obtained.

Complete step by step solution: As the given lines are z- axis and the line $x + y + 2z - 3 = 0$, $2x + 3y + z - 4 = 0$
By trial and error method first, calculate any two points that lie on the above line
So for the line ${\text{x + y + 2z - 3 = 0}}$ the points is $(1, - 2,2)$ this is obtained by randomly putting the number in equation and checking it.
And for the line, $2x + 3y + z - 4 = 0$, points are $(5, - 2,0)$.
Now, write the equation of line using both the obtained points as $(1, - 2,2)$ and $(5, - 2,0)$.
So, the equation of line is $\dfrac{{x - 5}}{4} = \dfrac{{y + 2}}{0} = \dfrac{z}{{ - 2}}$,
And the equation of z- axis is $\dfrac{x}{0} = \dfrac{y}{0} = \dfrac{z}{1}$.
Now we know shortest distance between two line is given by, $d = \dfrac{{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )}}{{\left| {(\overrightarrow b \times \overrightarrow d )} \right|}}$
So, from all the above equation we can write the following vectors as,

$$\overrightarrow a = 0 \\\ \overrightarrow b = \widehat k \\\ \overrightarrow c = 5\widehat i - 2\widehat j \\\ \overrightarrow d = 4\widehat i - 2\widehat k \\\$$

Hence, putting all the above vectors in above formula,
$d = \dfrac{{(0 - 5\widehat i + 2\widehat j).(\widehat k \times (4\widehat i - 2\widehat k))}}{{\left| {\widehat k \times (4\widehat i - 2\widehat k)} \right|}}$
Now, calculating the above cross-product using determinant method so $\widehat k \times (4\widehat i - 2\widehat k)$ can be given as

{\widehat i}&{\widehat j}&{\widehat k} \\\ 0&0&1 \\\ 4&0&{( - 2)} \end{array}} \right|$$ Hence on expanding above determinant, the above vector can be given as

= - \widehat j(0 - 4) \\
= 4\widehat j \\

Now, putting this in the above distance formula, $$d = \dfrac{{(0 - 5\widehat i + 2\widehat j).(4\widehat j)}}{{\left| {4\widehat j} \right|}}$$………….(1) Now, calculating the dot product of the numerator So it can be simplified to

( - 5\widehat i + 2\widehat j).(4\widehat j) = 8{(\widehat j)^2} = 8 \\
\because {(\widehat j)^2} = 1 \\

As, $$\left| {4\widehat j} \right| = 4$$ Hence, (1) can be written as,

\therefore d = \dfrac{{(8)}}{{\left| {4\widehat j} \right|}} \\
\therefore d = \dfrac{{(8)}}{4} \\
\therefore d = 2 \\

$$**Hence, option(b) is correct answer.** **Note:** In geometry, we often deal with different sets of lines such as parallel lines, intersecting lines or skew lines. The distance is the perpendicular distance from any point on one line to the other line. Just for a better understanding, we should know that perpendicular distance gives is the shortest distance.$$